Electromagnetism I Chapter I Prof. Dr T. Fahmy
CHAPTER I
THE ELECTRIC FIELD AND POTENTIAL DIFFERENCE
After completing this chapter, the students should know:
· The different types of electric charges.
· The concept of Coulomb’s law.
· The relation between electric force and electric field.
· The units of electric charge, electric force and electric field.
· The concept of Gauss’s law.
· The relation between potential energy and potential difference.
· The motion of charged particle.
CHAPTER I
THE ELECTRIC FIELD AND POTENTIAL DIFFERENCE
The Electrical Force
Coulomb’s Law:
Coulomb measured the magnitudes of the electric forces between charged objects using the torsion balance, and he showed that:
1- The electric force is inversely proportional to the square of the separation r between the particles and directed along the line joining them, i.e.
2- The electric force is directly proportional to the product of the charges q1 and q2 on the particles.
Therefore, from these observations, we can express Coulomb’s law as follow:
Where kc is a constant called the Coulomb’s constant. The value of the Coulomb’s constant depends on the choice of units. The Coulomb’s constant kc in SI units has the value of
Where e0 is known as the permittivity of free space and has the value of 8.85x10-12 C2/N m2. The smallest unit of charge known in nature is the charge on the electron or proton, which has an absolute value of
Note that:
The electric force is attractive if the charges are of opposite sign and repulsive if the charges have the same sign.
Example (1.1): (HW)
What is the total charge of 75.0 kg of electrons?
Solution:
The mass of one electron is 9.11 × 10−31 kg, so that a mass M = 75.0 kg contains
N = M/me = (75.0 kg)/(9.11 × 10−31 kg)
= 8.23 × 1031 electrons
The charge of one electron is −e = −1.60 × 10−19 C, so that the total charge of N
electrons is:
Q = N(−e) = (8.23 × 1031)(−1.60 × 10−19 C) = −1.32 × 1013 C
Example (1.2): (HW)
A point charge of +3.00 × 10−6 C is 12.0 cm distant from a second point charge of −1.50 × 10−6 C. Calculate the magnitude of the force on each charge.
Solution:
Being of opposite signs, the two charges attract one another, and the magnitude of this
force is given by Coulomb’s law,
Each charge experiences a force of attraction of magnitude 2.81N.
Example (1.3):
What must be the distance between point charge q1 = 26.0 μC and point charge q2 = −47.0 μC for the electrostatic force between them to have a magnitude of 5.7 N?
Solution:
According to Coulomb’s law, we have an attraction force
Example (1.4)
Two small positively charged spheres have a combined charge of 5.0 × 10−5 C. If each sphere is repelled from the other by an electrostatic force of 1.0N when the spheres are 2.0m apart, what is the charge on each sphere?
Solution:
The condition on the combined charge of the spheres gives us:
Q1 + Q2 = 5 × 10−5 C (1)
the next condition gives us:
(2)
From equation 1, we get:
Q2 = Q1 - 5 × 10−5
Also from equation (2), we get:
Q1(5 × 10−5 − Q1) = 4.45× 10−10
Q21 − 5 × 10−5xQ1 + 4.45× 10−10 = 0
Then:
Q1 = 3.84 × 10−5 C → Q2 = 5.0 × 10−5 C − Q1 = 1.16 × 10−5 C
Q1 = 1.16 × 10−5 C → Q2 = 5.0 × 10−5 C – Q1 = 3.84 × 10−5 C
Example (1.5):
Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.108N when separated by 50.0 cm, center-to-center. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.360N. What were the initial charges on the spheres?
Solution:
If the force of attraction between them has magnitude 0.108N, then Coulomb’s law gives us:
But since we are told that the charges attract one another, we know that q1 and q2 have opposite signs and so their product must be negative. So we can drop the absolute value sign if we write:
Then the two spheres are joined by a wire. If the new charge on each sphere is Q, then:
The force of repulsion between the spheres is now 0.0360N, so that:
We don’t know what the sign of Q is, so we can only say:
Also, note that:
Then:
Hence:
The solution is:
Also
Example (1.6):
The electron and proton of a hydrogen atom are separated by a distance of approximately 5.3x10-11 m. Find the magnitudes of the electric force and the gravitational force between the two particles, knowing that, me= 9.11x10-31 kg, mp= 1.67x10-27, kc= 9x109 N.m2/m2 and G= 6.7x10-11 N.m2/kg2.
Solution:
Firstly, the Coulomb force is
Secondly, the gravitational force is
Example (1.7):
Two charges, one of is +8x10−6 C and the other is −5x10−6 C, attract each other with a force of −40 N. How far apart are they?
Solution:
According to Coulomb’s law, we have
Example (1.8):
Consider three point charges located at the corners of a right triangle as shown in figure, where q1=q3= 5 mC and q2 = -2 mC and a= 0.1 m.
Find the resultant force exerted on q3.
Solution:
Note that, the force F23 exerted
by q2 on q3 is attractive
Therefore, the magnitude of F23 is
Also, the force F13 exerted by q1 on q3 is repulsive
Therefore, the magnitude of F13 is
But F13 must be analyzed in two components in x-axis (horizontal component) and y-axis (vertical component), as follow:
Therefore, the total force in x-axis is
Fx= 7.9 – 9 = -1.1 N
and the total force in y-axis is
Fy= 7.9 N
Then, the total force acting on the charge q3 is
Example (1.9):
Consider three point charges located at the corners of a right triangle as shown in figure, where q1=2 mC, q2= 5 mC and q3 = 4 mC. Find the resultant force exerted on q3.
Solution:
Note that, the force F13 exerted by q1 on q3 is repulsive
Therefore, the magnitude of F13 is
Also, the force F23 exerted by q2 on q3 is repulsive
Therefore, the magnitude of F23 is
But F23 must be analyzed in two components in x-axis (horizontal component) and y-axis (vertical component), as follow:
Therefore, the total force in x-axis is
Fx= 43.2 + 80 = 123.2 N
and the total force in y-axis is
Fy= 57.6 N
Then, the total force acting on the charge q3 is
The Electric Field
The electric field E at a point in space is defined as ‘ the electric force F acting on a positive test charge q0 placed at that point divided by the magnitude of the test charge’’
Therefore,
To define the direction of an electric field, consider a point charge q located at a distance r from a test charge q0 located at a point p.
According to Coulomb’s law, the electric force exerted
by q on q0 is
Also, note that:
At any point p the total electric field due to a group of charges equals the vector sum of the electric fields of the individual charges.
Then:
Rules for drawing electric field lines:
· Lines begin at positive charges and end at negative charges or infinity.
· The number of lines drawn leaving (approaching) a positive (negative) charge is proportional to the magnitude of the charge.
· No two field lines can cross each other.
Example:
Example (1.10):
Two point charges have the same charge q with opposite sign located at a distance r between them. Calculate the electric force and electric field on a third charge q3 if this charge is located at different positions such as a, b, c and d. Knowing that,
q1 and q2 = 5 mC, q3 = 4 mC and r= 0.12 m.
Solution:
At the point (a):
At the point (b):
There are two forces against to each other, therefore, we will obtain the total force as a difference between them as follow
At the point (c):
There are two forces against to each other, therefore, we will obtain the total force as a difference between them as follow
At the point (d)
Example (1.11):
The electric field (E) in a certain neon sign is 5000 V/m. What force does this field (F) exert on a neon ion of mass 3.3x10−26 kg and charge +e?
Solution:
The force (F) on the neon ion is
Example (1.12):
In the following figure calculate
1- The total force () acting on Q2, due to the effect Q1 and Q3
2- The electric field () at the position of Q2.
Solution:
1- The total force is
= +
= =
= (to the right)
= =
= (to the left)
Therefore, the total force is
= += 180 - 50 =130 N (to the right)
2- The electric field is
= = 3.25x107 N/m
Example (1.13):
In the following figure calculate
3- The total force () acting on Q2, due to the effect Q1 and Q3
4- The electric field () at the position of Q2.
Solution:
3- The total force is
= + (to the left)
= =
=
= =
=
Therefore, the total force is
= += 180 + 200 =380 N
4- The electric field is
= = 5x107 N/m
Example (1.14):
Find the force on the charge q2 in the diagram below due to the charges q1 and q3.
Solution:
Example (1.15):
Find the force on q2 in the diagram to the right.
Example (1.16):
Find the electric field at P due to charges q1 and q2.
Example (1.17):
A charge q1 = 7.0 mC is located at the origin, and a second charge q2 = -5.0 mC is located on the x axis, 0.30 m from the origin. Find the electric field at the point P, which has coordinates (0, 0.40) m. /Solution:
First, let us find the magnitude of the electric field at P due to each charge. The fields E1 due to the 7.0 mC charge and E2 due to the -5.0 mC charge are:
The vector E1 has only a y component. The vector E2 has an x component given by
E2 cos θ =3/5 E2 and a negative y component given by -E2 sin θ =- 4/5 E2. Hence, we can express the vectors as
The resultant field E at P is the superposition of E1 and E2:
From this result, we find that E makes an angle φ of 66° with the positive x axis and has a magnitude of 2.7 x105 N/C.
Gauss's Law
This law is relating the distribution of electric charge to the resulting electric field. Gauss's law states that:
The electric flux through any closed surface is proportional to the enclosed electric charge.
The law was formulated by C. F. Gauss in 1835, but was not published until 1867. It is one of four of Maxwell's equations which form the basis of classical electrodynamics, the other three being Gauss's law for magnetism, Faraday's law of induction, and Ampère's law with Maxwell's correction. Gauss's law can be used to derive Coulomb's law, and vice versa.
As shown in the figure, if a surface with an
Area of A is placed in an electric field of intensity
E, therefore, the total normal electric flux can be expressed (ɸ) as follow:
(1.8)
So, the vertical flux (dɸ) through an area of (dS) can be
written as follow:
(1.9)
where E is the electric field at any point of sphere surface and equals to
Then by using the integration, the total flux can be determined:
(1.10)
But if the surface of the sphere is non-homogeneous, the flux will be
Or in other form as follow:
(1.12)
Therefore, from equations (1.11) and (1.12), we can conclude that:
(1.13)
Example (1.18):
Calculate the electric flux (φ) through a circle has a radius of 5 cm and a charge of 4 mC at its center.
Solution:
φ= E. A
r= 5x10-2 m and Q=4x10-6 C
The Potential Energy (U) and Potential Difference (V)
If a test charge q0 is placed in an electric field E, created by some other charged objects, the electric force (F) acting on the test charge is q0 E. If the charge is moved a distance d , then the total work done can be written as follow:
Where θ is the angle between the force (F) and the displacement (d). Therefore if the charge is moved from the point A to the point B, the total work done can be written as follow:
Also, in terms of the potential energy (U), the equation (1.15) can be written as follow
Where
Then
The negative sign mean that, the motion against the electric field direction.
Note that,
· at θ= 00 → WAB and UAB have a maximum value
· at θ= 900 → WAB and UAB have a minimum value, i.e., (WAB and UAB = 0).
The Potential Difference (V)
The potential difference (V) is defined as the work done to transport a positive charge against the electric field.
Also,
The potential difference (V) is defined as the ratio between the potential energy (U) to a test charge (q0).